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Question Number 202154 by cortano12 last updated on 22/Dec/23

⇃

$$\:\:\:\downharpoonleft\underline{\:} \\ $$

Commented by mr W last updated on 22/Dec/23

f(xy+1)= 𝛎((x)f(y)−2f(y)−x+5 )

$$\mathrm{f}\left(\mathrm{xy}+\mathrm{1}\right)=\:\:\boldsymbol{\nu} \\ $$

Answered by Rasheed.Sindhi last updated on 22/Dec/23

⇃

$$\:\:\:\downharpoonleft\underline{\:} \\ $$

Commented by Rasheed.Sindhi last updated on 22/Dec/23

f(xy+1)= J((x)f(y)−2f(y)−x+5 )

$$\:\:\mathrm{f}\left(\mathrm{xy}+\mathrm{1}\right)=\:\:\mathcal{J} \\ $$

Answered by cortano12 last updated on 22/Dec/23

{ ((f(xy+1)=f(x)f(y)−2f(y)−x+5)),((f(yx+1)=f(y)f(x)−2f(x)−y+5)) :} ⇒(1)−(2) : 0=−2f(y)+2f(x)+y−x 2f(y)−2f(x)=y−x y=0⇒16−2f(x)=−x ⇒f(x)= ((16+x)/2) x=0⇒2f(y)= y+16 ⇒f(y)=((y+16)/2) f(2024)= ((2040)/2)=1020

$$\:\begin{cases}{\mathrm{f}\left(\mathrm{xy}+\mathrm{1}\right)=\mathrm{f}\left(\mathrm{x}\right)\mathrm{f}\left(\mathrm{y}\right)−\mathrm{2f}\left(\mathrm{y}\right)−\mathrm{x}+\mathrm{5}}\\{\mathrm{f}\left(\mathrm{yx}+\mathrm{1}\right)=\mathrm{f}\left(\mathrm{y}\right)\mathrm{f}\left(\mathrm{x}\right)−\mathrm{2f}\left(\mathrm{x}\right)−\mathrm{y}+\mathrm{5}}\end{cases} \\ $$$$\:\:\Rightarrow\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\::\:\mathrm{0}=−\mathrm{2f}\left(\mathrm{y}\right)+\mathrm{2f}\left(\mathrm{x}\right)+\mathrm{y}−\mathrm{x} \\ $$$$\:\:\mathrm{2f}\left(\mathrm{y}\right)−\mathrm{2f}\left(\mathrm{x}\right)=\mathrm{y}−\mathrm{x} \\ $$$$\:\:\mathrm{y}=\mathrm{0}\Rightarrow\mathrm{16}−\mathrm{2f}\left(\mathrm{x}\right)=−\mathrm{x} \\ $$$$\:\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\:\frac{\mathrm{16}+\mathrm{x}}{\mathrm{2}}\: \\ $$$$\:\:\mathrm{x}=\mathrm{0}\Rightarrow\mathrm{2f}\left(\mathrm{y}\right)=\:\mathrm{y}+\mathrm{16} \\ $$$$\:\:\Rightarrow\mathrm{f}\left(\mathrm{y}\right)=\frac{\mathrm{y}+\mathrm{16}}{\mathrm{2}} \\ $$$$\:\mathrm{f}\left(\mathrm{2024}\right)=\:\frac{\mathrm{2040}}{\mathrm{2}}=\mathrm{1020}\: \\ $$

Commented by Rasheed.Sindhi last updated on 22/Dec/23

Cortano sir! f(x)= ((16+x)/2) ⇒f(1)=((17)/2) whereas for x=y=0: f(xy+1)=f(x)f(y)−2f(y)−x+5 ∧ f(0)=8 ⇒f(0×0+1)=f(0)f(0)−2f(0)−0+5 ⇒f(1)=8^2 −2(8)+5=53 How f(1) has two different values?

$$\mathrm{Cortano}\:\mathrm{sir}! \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\:\frac{\mathrm{16}+\mathrm{x}}{\mathrm{2}}\:\Rightarrow\mathrm{f}\left(\mathrm{1}\right)=\frac{\mathrm{17}}{\mathrm{2}} \\ $$$$\mathrm{whereas} \\ $$$$\mathrm{for}\:\mathrm{x}=\mathrm{y}=\mathrm{0}: \\ $$$$\mathrm{f}\left(\mathrm{xy}+\mathrm{1}\right)=\mathrm{f}\left(\mathrm{x}\right)\mathrm{f}\left(\mathrm{y}\right)−\mathrm{2f}\left(\mathrm{y}\right)−\mathrm{x}+\mathrm{5}\:\wedge\:\mathrm{f}\left(\mathrm{0}\right)=\mathrm{8} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{0}×\mathrm{0}+\mathrm{1}\right)=\mathrm{f}\left(\mathrm{0}\right)\mathrm{f}\left(\mathrm{0}\right)−\mathrm{2f}\left(\mathrm{0}\right)−\mathrm{0}+\mathrm{5} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{1}\right)=\mathrm{8}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{8}\right)+\mathrm{5}=\mathrm{53} \\ $$$$\mathrm{How}\:\mathrm{f}\left(\mathrm{1}\right)\:\mathrm{has}\:\mathrm{two}\:\mathrm{different}\:\mathrm{values}? \\ $$

Commented by cortano12 last updated on 22/Dec/23

yes sir, the question inconsistent

$$\mathrm{yes}\:\mathrm{sir},\:\mathrm{the}\:\mathrm{question}\:\mathrm{inconsistent} \\ $$

Answered by MathematicalUser2357 last updated on 06/Jan/24

1020

$$\mathrm{1020} \\ $$

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